Difference between revisions of "12-240/Classnotes for Tuesday September 18"

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(Complex number)
(Complex number)
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In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎
 
In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎
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== Complex number==
 
== Complex number==
  
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Proof (1): Show that each of the field axioms hold for '''C'''.
 
Proof (1): Show that each of the field axioms hold for '''C'''.
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 +
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Ex. F1(a): ƶ1 + ƶ2 = ƶ2 + ƶ1, where ƶ1 = (a1, b1) and ƶ2 = (a2, b2)
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LHS: (a1,b1)+(a2,b2) = (a1+a2, b1+b2)
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RHS: (a2,b2)+(a1,b1) = (a2+a1, b2+b1)
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LHS=RHS by F1 of '''R'''
  
 
== Lecture 3, scanned notes upload by [[User:Starash|Starash]] ==
 
== Lecture 3, scanned notes upload by [[User:Starash|Starash]] ==

Revision as of 22:42, 18 September 2012

Various properties of fields

Thrm 1: In a field F: 1. a+b = c+b ⇒ a=c

2. b≠0, a∙b=c∙b ⇒ a=c

3. 0 is unique.

4. 1 is unique.

5. -a is unique.

6. a^-1 is unique (a≠0)

7. -(-a)=a

8. (a^-1)^-1 =a

9. a∙0=0 **Surprisingly difficult, required distributivity.

10. ∄ 0^-1, aka, ∄ b∈F s.t 0∙b=1

11. (-a)∙(-b)=a∙b

12. a∙b=0 iff a=0 or b=0

. . .

16. (a+b)∙(a-b)= a^2 - b^2 [Define a^2 = a∙a] Hint: Use distributive law


Thrm 2: Given a field F, there exists a map Ɩ: Z → F with the properties (∀ m,n ∈ Z):

1) Ɩ(0) =0, Ɩ(1)=1

2) Ɩ(m+n) = Ɩ(m) +Ɩ(n)

3) Ɩ(mn) = Ɩ(m)∙Ɩ(n)

Furthermore, Ɩ is unique.

Rough proof:

Test somes cases:

Ɩ(2) = Ɩ(1+1) = Ɩ(1) + Ɩ(1) = 1 + 1 ≠ 2

Ɩ(3) = Ɩ(2 +1)= Ɩ(2) + Ɩ(1) = 1+ 1+ 1 ≠ 3

. . .

Ɩ(n) = 1 + ... + 1 (n times)

Ɩ(-3) = ?

Ɩ(-3 + 3) = Ɩ(-3) + Ɩ(3) ⇒ Ɩ(-3) = -Ɩ(3) = -(1+1+1)

What about uniqueness? Simply put, we had not choice in the definition of Ɩ. All followed from the given properties.

At this point, we will be lazy and simply denote Ɩ(3) = 3_f [3 with subscript f]


∃ m≠0, m ∈ N, Ɩ(m) =0

In which case, there is a smallest m<0, for which Ɩ(m)=0. 'm' is the characteristic of F. Denoted char(F). Examples: char(F_2)=2, char(F_3)=3... but NOTE: char(R)=0


Thrm: If F is a field and char(F) >0, then char(F) is a prime number.

Proof: Suppoer char(F) =m, m>0. Suppose also m is not prime: m=ts, t,s ∈ N.

Then, Ɩ(m) = 0 = Ɩ(t)∙Ɩ(s) ⇒ Ɩ(t)=0 or Ɩ(s)=0 by P12.

If Ɩ(t)=0 ⇒ t≧m ⇒ m=t, s=1 or likewise for Ɩ(s)=0, and m=s, t=1

In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎

Complex number

Abstraction, generalization, definition, examples, properties, dream, implications, realization = formalization, PROOF.

Consider that fact that in R, ∄ x s.t. x^2 = -1

Dream: Add new number element 𝒊 to R, so as to still get a field & 𝒊^2 = -1

Implications: By adding 𝒊, we must add 7𝒊, and 2+7𝒊, 3+4𝒊, (2+7𝒊)∙(3+4𝒊), (2+7𝒊)^-1, etc.

So, how do we define this field?

Definition

C = {(a,b): a,b ∈ R} Also, 0 (of the field) = (0,0); 1( of the field) = (1,0)

Define addition: (a,b)+(c,d) = (a+c, b+d)

Define multification: (a,b)(c,d) = (ad-bd, ad+bc)

Theorems:

Thrm. 1. (C, 0, 1, +, ∙) is a field.

Thrm. 2. ∃ 𝒊 ∈ C s.t. 𝒊^2 = -1

Thrm. 3. C contains R

Proof (1): Show that each of the field axioms hold for C.


Ex. F1(a): ƶ1 + ƶ2 = ƶ2 + ƶ1, where ƶ1 = (a1, b1) and ƶ2 = (a2, b2) LHS: (a1,b1)+(a2,b2) = (a1+a2, b1+b2) RHS: (a2,b2)+(a1,b1) = (a2+a1, b2+b1) LHS=RHS by F1 of R

Lecture 3, scanned notes upload by Starash